Q.

In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s and 2.80 s. Calculate the mean absolute error.

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a

0.11 s

b

0.92 s 

c

0.42 s

d

0.10 s

answer is A.

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Detailed Solution

The mean period of oscillation of the pendulum,
Tmean=2.63+2.56+2.42+2.71+2.805=13.125=2.624=2.62s
The absolute errors in the measurements are
T1 = 2.63 s - 2.62 s = 0.01 s 
T2 = 2.56 s - 2.62 s = -0.06 s 
T3 = 2.42 s - 2.62 s = -0.20 s 
T4 = 2.71 s - 2.62 s = 0.09 s
T5 = 2.80 s - 2.62 s = 0.18 s
The arithmetic mean of all the absolute errors is
Tmean=i=15Ti5
= [(0.01 + 0.06 + 0.20 + 0.09 + 0.18)] / 5
= 0.54 5/ = 0.108  0.11 s

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